Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1000000000000000000LL;
const int MAXN = 100005;
const int MAXLG = 17;
int n, m, q;
vector<int> T[MAXN];
vector<int> res[MAXN];
int dep[MAXN];
int dad[MAXN];
int jmp[MAXN][MAXLG];
vector<int> mins[MAXN][MAXLG];
void dfs(int i, int d) {
  dep[i] = d;
  for (auto j : T[i]) {
    if (j == dad[i]) continue;
    dad[j] = i;
    dfs(j, d + 1);
  }
}
int lca(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  for (int lg = MAXLG - 1; lg >= 0; --lg)
    if (dep[jmp[a][lg]] >= dep[b]) a = jmp[a][lg];
  if (a == b) return a;
  for (int lg = MAXLG - 1; lg >= 0; --lg) {
    if (jmp[a][lg] != jmp[b][lg]) {
      a = jmp[a][lg];
      b = jmp[b][lg];
    }
  }
  return dad[a];
}
void get_mins(int a, int pa, vector<int> &sol) {
  int delta = dep[a] - dep[pa] + 1;
  for (int lg = MAXLG - 1; lg >= 0; --lg) {
    if (((delta >> lg) & 1) == 1) {
      for (int x : mins[a][lg]) sol.push_back(x);
      a = jmp[a][lg];
    }
  }
}
void query(int a, int b, vector<int> &sol) {
  int c = lca(a, b);
  get_mins(a, c, sol);
  get_mins(b, c, sol);
  sort(sol.begin(), sol.end());
  sol.resize(unique(sol.begin(), sol.end()) - sol.begin());
  while (sol.size() > 10) sol.pop_back();
}
int main(void) {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    T[a].push_back(b);
    T[b].push_back(a);
  }
  for (int i = (0); i < (m); ++i) {
    int c;
    scanf("%d", &c);
    res[c].push_back(i);
  }
  for (int i = (1); i < (n + 1); ++i) {
    sort(res[i].begin(), res[i].end());
    while (res[i].size() > 10) res[i].pop_back();
  }
  T[0].push_back(1);
  dfs(0, 0);
  for (int i = (0); i < (n + 1); ++i) {
    jmp[i][0] = dad[i];
    mins[i][0] = res[i];
  }
  for (int lg = 1; lg < MAXLG; ++lg) {
    for (int i = (0); i < (n + 1); ++i) {
      int j = jmp[i][lg - 1];
      jmp[i][lg] = jmp[j][lg - 1];
      mins[i][lg] = mins[i][lg - 1];
      for (int x : mins[j][lg - 1]) mins[i][lg].push_back(x);
      sort(mins[i][lg].begin(), mins[i][lg].end());
      while (mins[i][lg].size() > 10) mins[i][lg].pop_back();
    }
  }
  vector<int> ans;
  ans.reserve(10);
  for (int i = (0); i < (q); ++i) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    ans.clear();
    query(u, v, ans);
    while ((int)ans.size() > a) ans.pop_back();
    printf("%d ", (int)ans.size());
    for (int j = (0); j < ((int)ans.size()); ++j)
      printf("%d%c", ans[j] + 1, " \n"[j + 1 == (int)ans.size()]);
    if (ans.empty()) puts("");
  }
  return 0;
}